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I posted recently trying to work out how many combinations were possible for my scheduling algorithm of teachers to classes. So if I had 4 classes and 3 teachers The combinations would be 4^3 right? I forgot to mention that not all classes need to be filled. So if I had 1 teacher and 3 classes the possibilities would be. The one teacher takes these classes "takes none" -1 - 2 - 3 - 1,2 - 1,3- 1,2,3 -2,3 So there are 8 possibilities for one teacher and 3 classes

What would be the maths for this on a larger scale, such as 50 classes, ten teachers Thankyou

Aleddd
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  • Is it the case that every teacher teaches 1 class (some don't get taught) or that every class gets 1 teacher (some teachers get more than one class)? – TravisJ Apr 18 '15 at 01:27
  • some will not get taught. – Aleddd Apr 18 '15 at 01:28
  • no lessons take place // lesson 1 takes place, 2 and 3 canceled // lesson 2 takes place 1 and 3 cancelled and so on – Aleddd Apr 18 '15 at 01:29
  • So do you have the condition that: each of $n$ classes can be assigned at most one of $m$ teachers, but each teacher can teach multiple classes? – Graham Kemp Apr 18 '15 at 01:32
  • a teacher can teach as many classes. so all 3 // each class only has one teacher – Aleddd Apr 18 '15 at 01:33

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So, each of $n$ classes can be assigned at most one of $m$ teachers, but each teacher can teach multiple classes.

Thus there is choice of one of $m+1$ options, made $n$ times, with repetition allowed.

The count of distinct ways to do this is: $(m+1)^n$

EG: With one teacher and three classes we have $(1+1)^3=8$ ways.

With ten teachers, and fifty classes, we have $11^{50}$

Graham Kemp
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  • Great thank you

    didnt realise how big my search area was going to be for this scheduling algorithm

     – Aleddd Apr 18 '15 at 02:08